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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: interminable on March 26, 2010, 02:50:56 PM

Title: Synthesis of Copper (II) Acetate Question
Post by: interminable on March 26, 2010, 02:50:56 PM

In lab we synthesized copper (II) acetate starting from copper (II) sulfate pentahydrate. First ammonia was added, followed by sodium hydroxide.

I've been working on the balanced equations, and I understand we're forming a Cu(OH)2 precipitate twice, (we dissolved it with excess ammonia the first time and reformed it with the NaOH).

We dissolve the Cu(OH)2 in 10% acetic acid solution to obtain the acetate product. I'm just wondering why dissolving the copper (II) sulfate directly in the acetic acid and skipping all the other steps wouldn't work.

Title: Re: Synthesis of Copper (II) Acetate Question
Post by: Schrödinger on March 27, 2010, 03:32:40 AM

If you were to add acetic acid directly, then the reaction (supposing that they react) would be :

CuSO₄ + CH₃COOH  :rarrow: (CH₃COO)₂Cu + H₂SO₄

But as you can see, a weak acid (acetic acid) is displacing a stronger acid (sulphuric acid) from its salt. That's not possible.

On the other hand, when you add NaOH, the acid being displaced is H₂O, which is certainly weaker than acetic acid. (wirte the equation)


P.S : I recently learned on this forum that there might be rare exceptions to this rule. http://www.chemicalforums.com/index.php?topic=40518.0

Title: Re: Synthesis of Copper (II) Acetate Question
Post by: interminable on March 27, 2010, 12:35:32 PM

Thanks, the answer seems obvious now, haha.
 
Cu(OH)2 + CH3COOH --> Cu(CH3COO)2 + H2O

This is what you were referring to, right?

Title: Re: Synthesis of Copper (II) Acetate Question
Post by: Schrödinger on March 28, 2010, 08:30:21 AM

Yes